Jan, Problem 1 | Promotion Counting

Promotion Counting

Problem Statement

Given a tree, for each node find the number of its children who’s value is greater than the original node’s value.


If we iterate over the nodes with a DFS, we guarantee that we’ll visit all of the node’s children. Furthermore, we can guarantee that between entering and exiting a node, the only nodes processed are that node’s children.

More specifically, we maintain a BIT to keep track of our state. When we first visit a node, we sum from i - N. Then, we visit all of the node’s children. After visiting all of the children recursively, we again compute the sum of i - N. The difference is our answer. Finally, we update the BIT with our current node’s value and return.


My solution can be found here

Written by Robert Chen

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